Suppose we have a supported beam with a uniformly distributed load. The deflection $($y$)$ of the beam at any point x along its length can be approximated by the following equation:

$()=(+)$

Where Wo is the uniformly distributed load $($force per unit length$),$ E is Young$$s modulus of the material, I is the moment of inertia of the beam$$s cross$-$sectional area, L is the length of the beam, and x is the distance along the beam.

a$.$ Write a MATLAB function to calculate the deflection for value$($s$)$ of x$.$ Input in the code Wo$=2\mathrm{.}5$ KN$/$cm$,$ E$=50,000$ KN$/$cm$2,$ I$=30,000$ cm$4,$ and L$=600$ cm$.$ Enable the code to:

$$ calculate the deflection

$$ determine the maximum deflection and maximum deflection locations using MATLAB$$s

built$-$in function $$min$.$ The $$min$$ function in MATLAB returns the minimum value and the index where that minimum value occurs in the input vector $($for more information help min command$).$

syntax: $[$min$\_$value, index$]=$min$($variable name of the array containing all values$)$

Note that deflection values are negative numbers, and the max deflection $=$min value.

$$ use the index $($from the above output$)$ to identify the location of maximum deflection of

vector x$.$

Example vector x contains $5$ data points, then write a code to find the data that is stored in row $2($in this case index$=2)$

o x$=[2,3,5,7,9]$; then x$(2)=3$

$$ plot the deflection y$($x$)$ along the length of the beam

$$ mark the point of maximum deflection on the plot